Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in the summer of , Google's Chrome browser exceeded a market share for the first time, with a share of the browser market (Forbes website, December ). For a randomly selected group of Internet browser users, the following questions. A. Compute the probability that exactly of the Internet browser users use Chrome as their Internet browser.
B. Compute the probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.
C. For the sample of 20 Internet browser users, compute the expected number of Chrome users.
d. For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

question:

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step-by-step explanation:

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step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.